The Proof is 'not-so' Trivial - Physics Edition

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Yep sorry haha spoilers. I didn't actually make the question and also not my solution however I do agree with their solution...

Is that 0.685 radians or degrees? The ^c is a little confusing...

Your approach makes sense, and is the way I did it before looking at his solution, however it yields a different answer

This is their solution for the first bit:

Spoiler

Radians. That's the approach I initially tried but I didn't know that trig identity so got stuck and looked for another way.

Mine doesn't work (you can tell that angle's too low because it must be greater than 45 degrees). You can draw parabolas that have maximum points at (x,2) where x is greater than two. I've given the launch angle for one of those I think and so the pizza hits the wall just below the roof. :frown:

Reply 21

lerjj

Original post by Phichi

Solution 6

1). Will hit the walls during its descent, due to the rotation of the planet. If you bore through the axis of rotation, as suggested, this won't occur.

2).

Simple:

Spoiler

Will update with a more complex solution revolving more around polar co-ords soon.

Even simpler, although not very rigorous because my internet's breaking on me (making LaTeX rather difficult):

Spoiler

Reply 22

Phichi

Original post by lerjj

Even simpler, although not very rigorous because my internet's breaking on me (making LaTeX rather difficult):

Spoiler

Indeed, if the earth was hollow, with a thin shell surface, there'd be no net force on a mass inside. You're suggesting splitting the earth up into an infinite amount of thin spherical shells. As you fall the shells you've already past will contribute nothing to the force acting upon you, but the shells further inside the earth, ones you haven't past yet, will still exert a force as if the mass was concentrated at the center, given by

F = -\dfrac{GMm}{r^2}

. It's exactly the same idea, just a different way of thinking about it. But showing the net force from a shell you've already passed is not as easy as the first method, if you're trying to be rigorous. Show some working, I'd like to see it all.

(edited 9 years ago)

Reply 23

langlitz

Should I upload the solution for some of Problem 3?

Reply 24

lerjj

Original post by Phichi

F = -\dfrac{GMm}{r^2}

You don't need an infinite number of shells- compress all the mass you've passed into one thin subshell and all the mass further in than you acts like a point mass.

Spoiler

(edited 9 years ago)

Gravitational shell.png4.5KB

Reply 25

Phichi

Original post by lerjj

You don't need an infinite number of shells- compress all the mass you've passed into one thin subshell and all the mass further in than you acts like a point mass.

Spoiler

That's exactly what I said, the mass you've passed, its entirely irrelevant if you consider an infinite amount of shells, which is the only plausible way you could model it. And indeed, the shells you're yet to pass will act as a point mass, like I said, following the same force expression. You could say it was zero by intuition, but there is a way using geometry I think.

Reply 26

lerjj

Original post by Phichi

Ah, okay. I thought you were saying that the proof that it was zero was complicated. You could always form some complicated integral I guess and actually sum all of the attractions (I guess, my maths is worse than my physics...). I think I've seen a geometric proof for the inside of a hollow charged sphere which is the same situation... I'll try and find it.

Reply 27

Phichi

Original post by lerjj

I meant to say, the proof of that, to be rigorous, is more complex, then just the solution I posted, unless you just take assumptions. It's the same proof really though.

Reply 28

langlitz

Original post by lerjj

I believe you are referring to the shell theorem http://en.m.wikipedia.org/wiki/Shell_theorem

Posted from TSR Mobile

Reply 29

langlitz

Problem 7: *

Two, free to move, small spheres with charge q and 4q are placed a distance

d

apart. A third small charged sphere, also free to move, is placed so that all three charges remain in their positions. Find the location and charge of the third sphere.

(edited 9 years ago)

Reply 30

Phichi

Original post by langlitz

Problem 7: *

Two, free to move, small spheres with charge q and 4q are placed a distance

d

apart. A third small charged sphere, also free to move, is placed so that all three charges remain in their positions. Find the location and charge of the third sphere.

Is there more than one possible answer? Just by inspection. Will attempt later, didn't sleep last night, think I'm going to die.

Posted from TSR Mobile

Reply 31

Phichi

Problem 8*** (10 stars for LaTeX).

Given Planck's law is terms of frequency:

B_v(v, T) = \dfrac{2hv^3}{c^2}\dfrac{1}{exp \big( \dfrac{hv}{kt}\big) - 1}

1). Giving the steps involved, show that Planck's law in terms of wavelength is as follows:

B_v(\lambda, T) = \dfrac{2hc^2}{\lambda ^5}\dfrac{1}{exp \big( \dfrac{hc}{\lambda kt}\big) - 1}

2). By differentiating with respect to

\lambda

and equating to zero, using the substitution

x \equiv \dfrac{hc}{\lambda kT}

, show that the expression becomes

\dfrac{xe^x}{e^x-1} - 5 = 0

.

3). Using your own methods, prove Wien's displacement law:

\lambda _{max} = \dfrac{b}{T}

. Where

b \approx 2.898\times 10^{-3}\,\, m\cdot K

(edited 9 years ago)

Reply 32

Phichi

Original post by langlitz

Spoiler

Was hoping someone would answer this, thread is dying.

Firstly, we know the all the spheres need to be collinear. Thus, if you place the third charge some where inbetween the other two, we can solve this easily. Let the distance from charge 1 to charge 3 be r. We know

0 < r < d

Solution 7

The force on charge 1 given by charge 3, must be equal and opposite to the force given by charge 2 on charge 1.

Let the charge of the third sphere be Q.

Thus:

k_e \dfrac{qQ}{r^2} = -k_e \dfrac{4q^2}{d^2}

This simplifies to

Q = -\dfrac{4qr^2}{d^2}

Using the same logic for the two forces on charge 3:

k_e \dfrac{4q^2}{d^2} = -k_e \dfrac{4qQ}{(d-r)^2} \, \, \, \Rightarrow \, \, Q = -\dfrac{q(d-r)^2}{d^2}

Equating either equations.

\dfrac{4qr^2}{d^2} = \dfrac{q(d-r)^2}{d^2} \, \, \, \Rightarrow \, \, 4r^2 = (d-r)^2 \, \, \, \Rightarrow \, \, (3r-d)(r+d) = 0

Seeing as the charge lies between the other two charges,

r \neq d

. Thus

r = \dfrac{d}{3}

Subbing this in for the first equation, we get

Q = -\dfrac{4q}{9}

The third sphere lies

\dfrac{d}{3}

away from the sphere with charge q, in between the charge q and 4q. It is

\dfrac{2d}{3}

away from the 4q charge. The charge of the third sphere is

-\dfrac{4q}{9}

(edited 9 years ago)

Reply 33

Phichi

Original post by langlitz

Original post by lerjj

Original post by WishingChaff

Post up guys :smile:

Dying slowly.

(edited 9 years ago)

Reply 34

langlitz

Original post by Phichi

Equating either equations.

\dfrac{4qr^2}{d^2} = \dfrac{4q(d-r)^2}{d^2} \, \, \, \Rightarrow \, \, r^2 = (d-r)^2 \, \, \, \Rightarrow \, \, r = \dfrac{d}{2}

Subbing this in for the first equation, we get

Q = q

The third sphere is directly inbetween the two other spheres on a straight line, with charge q.

That's not correct, you can see just by looking at it, if all the charges are positive then they aren't going to stay in the same place

Reply 35

Phichi

Original post by langlitz

That's not correct, you can see just by looking at it, if all the charges are positive then they aren't going to stay in the same place

Fixed. I think I posted that at 4am, my brain must've been malfunctioning, just missed out the two minus' when equating, all sorted.

Reply 36

langlitz

Original post by Phichi

Fixed. I think I posted that at 4am, my brain must've been malfunctioning, just missed out the two minus' when equating, all sorted.

Still doesn't make sense though.. If it's directly between the two charges but one charge is 4x larger than the other then the forces aren't equal

Reply 37

Phichi

Original post by langlitz

Still doesn't make sense though.. If it's directly between the two charges but one charge is 4x larger than the other then the forces aren't equal

Give me two minutes and expel me from my Physics degree please. I used 4q for both charges, I think a common saying occurs here, RTFQ. You can see both expressions are

(4q)(4q) \equiv 16q^2

. I'll fix it now. I really shouldn't stay up so late and post.

(edited 9 years ago)

Reply 38

langlitz

Original post by Phichi

Give me two minutes and expel me from my Physics degree please. I used 4q for both charges, I think a common saying occurs here, RTFQ. You can see both expression are